poj 2080 Calendar

你的名字 2022-06-16 02:07 95阅读 0赞

Calendar

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;13575</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4839</td> 
  </tr> 
 </tbody> 
</table>

Description

A calendar is a system for measuring time, from hours and minutes, to months and days, and finally to years and centuries. The terms of hour, day, month, year and century are all units of time measurements of a calender system.  
According to the Gregorian calendar, which is the civil calendar in use today, years evenly divisible by 4 are leap years, with the exception of centurial years that are not evenly divisible by 400. Therefore, the years 1700, 1800, 1900 and 2100 are not leap years, but 1600, 2000, and 2400 are leap years.  
Given the number of days that have elapsed since January 1, 2000 A.D, your mission is to find the date and the day of the week.

Input

The input consists of lines each containing a positive integer, which is the number of days that have elapsed since January 1, 2000 A.D. The last line contains an integer −1, which should not be processed.  
You may assume that the resulting date won’t be after the year 9999.

Output

For each test case, output one line containing the date and the day of the week in the format of "YYYY-MM-DD DayOfWeek", where "DayOfWeek" must be one of "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday" and "Saturday".

Sample Input

1730
    1740
    1750
    1751
    -1

Sample Output

2004-09-26 Sunday
    2004-10-06 Wednesday
    2004-10-16 Saturday
    2004-10-17 Sunday

题目大意：

题目的意思就是，开始从2000年开始，任意给定一个天数，让你求出是哪一年的星期几最大不会超过9999年，例如给定1天那么一定就是2000年的1月1日，星期日，就是这样的一个题。

题目解析：

这是一道非常简单的模拟题，一种很常规的思路，用的是打表的方法，首先判断是不是闰年，那么这一年就有366天，否则这一年一共是365天，然后判断给定的天数，直接对7取余，因为一周有七天，这样就可以判断出给出的天数是星期几，按顺序，先把大于366天或者365天（看这一年是不是闰年），先化成年，然后剩下的天数，用同样的方法化成月，最后剩下的就是天，然后输出就可以得到正确的答案了，是不是完全都不用思考，直接就可以敲代码，最后把我写的代码献上。

代码：

#include<cstdio>
    #include<iostream>
    using namespace std;
    int main()
    {
        int n,year,month,day,t;
        int p[12]= {31,28,31,30,31,30,31,31,30,31,30,31};
        int r[12]= {31,29,31,30,31,30,31,31,30,31,30,31};
        int y[10000];
        for(int i=0; i<10000; i++)
        {
            if((i%4==0&&i%100!=0)||i%400==0) y[i]=366;
            else y[i]=365;
        }
        while(~scanf("%d",&n)&&n!=-1)
        {
            n++;
            t=n%7;
            for(int i=0; i<10000; i++)
            {
                if(n<=y[i])
                {
                    year=i;
                    break;
                }
                else n-=y[i];
            }
            if(!((year%4==0&&year%100!=0)||year%400==0))
            {
                for(int i=0; i<12; i++)
                {
                    if(n<=p[i])
                    {
                        month=i+1;
                        break;
                    }
                    n-=p[i];
                }
            }
            else
            {
                for(int i=0; i<12; i++)
                {
                    if(n<=r[i])
                    {
                        month=i+1;
                        break;
                    }
                    n-=r[i];
                }
            }
            day=n;
            if(t==3)
                printf("%d-%02d-%02d Monday\n",year+2000,month,day);
            else if(t==4)
                printf("%d-%02d-%02d Tuesday\n",year+2000,month,day);
            else if(t==5)
                printf("%d-%02d-%02d Wednesday\n",year+2000,month,day);
            else if(t==6)
                printf("%d-%02d-%02d Thursday\n",year+2000,month,day);
            else if(t==0)
                printf("%d-%02d-%02d Friday\n",year+2000,month,day);
            else if(t==1)
                printf("%d-%02d-%02d Saturday\n",year+2000,month,day);
            else if(t==2)
                printf("%d-%02d-%02d Sunday\n",year+2000,month,day);
    
        }
        return 0;
    }